# How do you evaluate # e^( (3 pi)/2 i) - e^( ( 11 pi)/12 i)# using trigonometric functions?

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To evaluate (e^{\left(\frac{3\pi}{2}i\right)} - e^{\left(\frac{11\pi}{12}i\right)}) using trigonometric functions, we can express each exponential term in trigonometric form.

Recall Euler's formula: (e^{ix} = \cos(x) + i\sin(x)).

So,

(e^{\left(\frac{3\pi}{2}i\right)} = \cos\left(\frac{3\pi}{2}\right) + i\sin\left(\frac{3\pi}{2}\right)).

(\cos\left(\frac{3\pi}{2}\right) = 0) and (\sin\left(\frac{3\pi}{2}\right) = -1), so (e^{\left(\frac{3\pi}{2}i\right)} = -i).

Similarly,

(e^{\left(\frac{11\pi}{12}i\right)} = \cos\left(\frac{11\pi}{12}\right) + i\sin\left(\frac{11\pi}{12}\right)).

Using the half-angle identities, (\cos\left(\frac{11\pi}{12}\right)) and (\sin\left(\frac{11\pi}{12}\right)) can be computed.

Now, we can substitute these values back into the expression (e^{\left(\frac{3\pi}{2}i\right)} - e^{\left(\frac{11\pi}{12}i\right)}):

((-i) - (\cos\left(\frac{11\pi}{12}\right) + i\sin\left(\frac{11\pi}{12}\right))).

Combining the real and imaginary parts, we get:

((-i) - \cos\left(\frac{11\pi}{12}\right) - i\sin\left(\frac{11\pi}{12}\right)).

Re-arranging terms, we have:

(-\cos\left(\frac{11\pi}{12}\right) - i\sin\left(\frac{11\pi}{12}\right) - i).

This is the final answer.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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